Student name: Student ID: Math 265 (Butler) Midterm III, 10 November 2011
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1 Student name: Student ID: Math 265 (Butler) Midterm III, November 2 This test is closed book and closed notes. No calculator is allowed for this test. For full credit show all of your work (legibly!). Each problem is worth points (a total of 5 points).. ewrite 3+3x f(x, y) dy dx + as a single integral of the form 3???? f(x, y) dy dx + f(x, y) dx dy. 2 3 x 2 f(x, y) dy dx If we sketch the region that the three integrals corresponds to, we see that we are bounded by the lines y, y 3, y 3 + 3x and y x 2. The last two we can rewrite as x (/3)y and x y + (we take the positive square root because we want the right hand side, i.e., positive x-values). It is now easy to set up the desired integral and we have 3 y+ (/3)y f(x, y) dx dy Σ
2 2. Find the centroid (x, y) of the homogenous region (i.e., density can be assumed to be ), bounded by the curve r sin θ (this is known as a cardiod). Hint: Use symmetry to simplify the problem. The following might be useful: sin 2 θ ( sin θ) 4 sin θ 7π, and cos 2θ, 2 ( sin θ) 5 sin θ 5π. ( sin θ) 3 sin θ 5π 4, By symmetry of the curve we automatically known that x. We have that y y da da. We now proceed to find these integrals. Since the curve is given to us in polar coordinates we will do our integration using polar coordinates (i.e., in cartesian coordinates the curve corresponds to (x 2 + y 2 + y) 2 x 2 + y 2 which is just ugh). We have that θ 2π and r sin θ. First we will do the denominator da sin θ For the numerator we have y da Therefore we have r dr 2 ( sin θ)2 2 r2 r r sin θ ( 2 sin θ + 2 sin2 θ ) sin 2θ) (3 4 sin θ 4 cos 2θ) ( 3 4 θ + cos θ sin θ y r 2 sin θ dr θ2π θ ( 3π 2 + ) ( + ) 3π 2. 3 r3 sin θ r sin θ r 3 ( sin θ)3 sin θ 5π 4 y da 5π/4 da 3π/2 π 2π 5 6. So the centroid of the cardiod is located at (, 5 6). (ala the hint).
3 3. Willy Wonka has come out with his latest opus, the MiFLLe sphere. The MiFLLe sphere is a ball which has a centimeter radius and is densest at the center of the ball and decreases in density towards the edges. Thus the ball Melts Fast and Lasts Long (MFLL or MiFLLe). Thanks to his revolutionary new process Wonka has been able to get the density to be (3 2q) grams per cubic centimeter where q is the distance to the center of the ball. Wonka has asked you to determine the exact weight of a MiFLLe sphere. You decide to give an answer to Wonka using the techniques of integration. In particular you place a MiFLLe sphere at the origin and make your units measurable in centimeters. To simplify matters you decide to only do the part of the MiFLLe sphere in the positive octant (i.e., x, y, z ), and multiply by. Set up (but do not evaluate!) three integrals which represent the total weight, one using Cartesian coordinates, one using Cylindrical coordinates and one using Spherical coordinates. (Each integral represents the same value; don t forget to multiply by eight.) For Cartesian coordinates the density function is (3 2 x 2 + y 2 + z 2 ) and we are integrating over the part of the sphere of radius in the positive octant. So x ; given an x we have that y will go from to the top a circle of radius which is x 2 ; given an x and a y we have that z will go from to the top of a sphere of radius which is x 2 y 2. Therefore the integral in Cartesian coordinates is x 2 x 2 y 2 (3 2 x 2 + y 2 + z 2 ) dz dy dx. For Cylindrical coordinates the density function is (3 2 r 2 + z 2 ) and we are integrating over the part of the sphere of radius in the positive octant. So θ π/2; given a θ we have that r will range from to (i.e., the edge of the circle of radius ); given r and θ we have that z will go from to the top of a sphere of radius which is r 2. Therefore the integral in Cylindrical coordinates is π/2 r 2 (3 2 r 2 + z 2 )r dz dr. For Spherical coordinates the density function is (3 2ρ) and we are integrating over the part of the sphere of radius in the positive octant. So θ π/2 and φ π/2 and ρ. Therefore the integral in Spherical coordinates is π/2 π/2 (3 2ρ)ρ 2 sin φ dρ dφ.
4 4. Carry out one of the three integrals from the previous problem to find the total weight of the MiFLLe sphere for Willy Wonka. Hint: It is the MiFLLe sphere. We have from the last problem π/2 π/ (3 2ρ)ρ 2 sin φ dρ dφ π/2 π/2 π/2 π/2 π/2 π/2 π/2 π/2 θπ/2 4θ θ 2π. ( cos φ) (3ρ 2 2ρ 3 ) sin φ dρ dφ (ρ 3 ρ 2 ρ4 ) sin φ dφ sin φ dφ φπ/2 φ So the MiFLLe ball weighs exactly 2π grams. ρ
5 5. Evaluate the integral u x + y and v y. 4 y y ye x+y dx dy by first making the change of variables One of the following might be useful: we w dw (w )e w + C, w 2 e w dw (w 2 2w + 2)e w + C, and w 3 e w dw (w 3 3w 2 + 6w 6)e w + C. Looking at the bounds we have y so that v. Similarly we have that the inside integral will go from x y to x 4 y or from x +y to x +y 4 so that x +y 4 or x + y 2 or u 2. These are some easy bounds and so we have 2 ve u J(u, v) du dv. So the last part for us to finish the change of variables is to determine the Jacobian. We need to solve for x and y in terms of u and v. We have that u 2 x + y x + v and so x u 2 v, and of course we already have y v. So now we are ready to compute the Jacobian. J(u, v) det Putting this in we have 4 y y x u x v y u y v ye x+y dx dy det 2u 2u 2 The antiderivative of ue u is (u )e u and so we have 2 2uve u du dv 2v dv 2 2uve u du dv. v u2 ue u du v 2 (u )e u e 2. v u
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